Solve the logarithmic equation. log 3 ( 3 x + 9 ) = log 3 5 + log 3 12 \log_3\left(3x+9\right)=\log_35+\log_312 lo g 3 ( 3 x + 9 ) = lo g 3 5 + lo g 3 12

In this problem, we're asked to solve the log equation. And here, the equation we're given is a log base 3 of 3x plus 9 is equal to log base 3 of 5 plus log base 3 of 12. Now the very first thing I notice about all of these logs is that they all have this same base of 3, which tells me that I can likely simplify this down into log base 3 on one side, log base 3 on the other, and just set what I'm taking the log of equal to each other. So let's go ahead and try to simplify this down into a single log of the same base on both sides. Now I don't have any work to do on this side. This is just log base 3 of 3x+9. And then on the other side, I see that I have this addition happening, which tells me that I can go ahead and use the product rule and turn this addition into the multiplication inside of a single log. So this is all equal to log base 3 of 5 times 12, which is just 60. So this becomes a log base 3 of 3x+9 is equal to the log base 3 of 60. Now since these logs have the same base, I can just take what I'm taking the log of and set them equal to each other. So let's go ahead and do that over here. We have 3x+9 is equal to 60. And now we just have a basic linear equation to solve. So I'm gonna go ahead and subtract 9 from both sides. That will cancel over here, leaving me with 3x is equal to 51. 60 minus 9 gives me that 51. And we have one final step here, which is to isolate x completely and divide both sides by 3. Now that will cancel out over here, leaving me with x is equal to 51 divided by 3 which is just 17. So my final answer is that x is equal to 17. Thanks for watching, and I'll see you in the next one.

6. Exponential & Logarithmic Functions

Solving Exponential and Logarithmic Equations

College Algebra

6. Exponential & Logarithmic Functions

Solving Exponential and Logarithmic Equations

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Multiple Choice

Solve the logarithmic equation.
$\log_3\left(3x+9\right)=\log_35+\log_312$

No Solution

Verified step by step guidance

Start by using the properties of logarithms to combine the right side of the equation. The equation is $ \log_3(3x+9) = \log_3 5 + \log_3 12 $. Use the property $ \log_b m + \log_b n = \log_b (m \cdot n) $ to combine the logs on the right side.

Combine the logs on the right side: $ \log_3 5 + \log_3 12 = \log_3 (5 \cdot 12) = \log_3 60 $. Now the equation is $ \log_3(3x+9) = \log_3 60 $.

Since the logarithms have the same base, you can set the arguments equal to each other: $ 3x + 9 = 60 $.

Solve the equation $ 3x + 9 = 60 $ by isolating $ x $. Subtract 9 from both sides to get $ 3x = 51 $.

Divide both sides by 3 to solve for $ x $: $ x = \frac{51}{3} $.

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Struggling with College Algebra?

Solve the logarithmic equation.log3(3x+9)=log35+log312\log_3\left(3x+9\right)=\log_35+\log_312log3(3x+9)=log35+log312

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Solving Exponential and Logarithmic Equations practice set

Solve the logarithmic equation.
$\log_3\left(3x+9\right)=\log_35+\log_312$